Yahtzee Strategy

Riddler Express

You are playing a one-turn game of Yahtzee in which the only goal is to maximize the score on this single turn. After the second of three rolls, your five dice show $4, 4, 4, 5, 5$. Keeping all five scores $25$ points for the full house. Rerolling the two $5$s aims at the $50$-point Yahtzee but risks a lowly three- or four-of-a-kind, scored as the sum of the five dice. What is the best strategy?

The Riddler, FiveThirtyEight, March 9, 2018(original post)

Solution

The benchmark. Standing pat scores a flat $25$ points. Any reroll is worthwhile only if its expected score exceeds $25$. The decision is whether the upside of a Yahtzee outweighs the cost of frequently landing on a hand worth less than $25$.

Reroll the two $5$s. Two fresh dice land in $6 \times 6 = 36$ equally likely ways. Sort the outcomes by the resulting hand.

• Both come up $4$: $1$ way out of $36$. Hand is five $4$s, scored as a Yahtzee, $50$ points.

• Both match (but not both $4$): $5$ ways (both $1$, both $2$, both $3$, both $5$, or both $6$). Hand is three $4$s plus a pair, a full house worth $25$.

• Otherwise: $30$ ways. Hand is three or four $4$s with two non-matching extras, scored as the sum of the five dice (the puzzle’s stated convention for "lowly three- or four-of-a-kind"). The two rerolled dice sum on average to $7$, and the three kept $4$s contribute $12$, so the average sum is $19$.

Expected score from rerolling the two $5$s, $$E_{\text{re 5s}} = \frac{1}{36}\!\cdot\!50 + \frac{5}{36}\!\cdot\!25 + \frac{30}{36}\!\cdot\!19 = \frac{50 + 125 + 570}{36} = \frac{745}{36} \approx 20.69.$$ That falls well short of the certain $25$.

Every other reroll is also worse than $25$. The same enumeration over each of the $31$ proper subsets of dice to reroll, with the scoring category chosen optimally after the third roll (Yahtzee, four-of-a-kind, full house, large or small straight, three-of-a-kind, or sum), gives the table below. The best reroll, "keep one $4$ and one $5$, reroll the other three", scores about $22.76$ points on average. Even that best alternative is short of $25$.

So the best policy is to keep the hand exactly as it is. $$\boxed{\,\text{Keep all five dice and score the full house for } 25 \text{ points.}\,}$$

The computation

Enumerate every possible reroll subset of the hand $(4,4,4,5,5)$, average the optimal Yahtzee scoring category over the $6^k$ outcomes for $k$ dice rerolled, and return the best.

1. For each subset of kept dice, list every outcome of the rerolled dice.

2. Score each resulting hand by the maximum of (Yahtzee $50$, large straight $40$, small straight $30$, full house $25$, three- or four-of-a-kind = sum, plain sum).

3. Average and report the best strategy.

from itertools import product, combinations

def score(d):
    s = sorted(d)
    cs = sorted([s.count(x) for x in set(s)], reverse=True)
    tot = sum(s)
    best = tot
    if cs[0] == 5: best = max(best, 50)
    if cs == [3, 2]: best = max(best, 25)
    uniq = sorted(set(s))
    if uniq == [1,2,3,4,5] or uniq == [2,3,4,5,6]:
        best = max(best, 40)
    for start in (1, 2, 3):
        if all(v in uniq for v in range(start, start + 4)):
            best = max(best, 30); break
    return best

hand = (4, 4, 4, 5, 5)
results = []
seen = set()
for r in range(6):
    for keep_idx in combinations(range(5), r):
        kept = tuple(sorted(hand[i] for i in keep_idx))
        if kept in seen: continue
        seen.add(kept)
        n_re = 5 - r
        if n_re == 0:
            ev = score(kept)
        else:
            tot = 0
            for out in product(range(1, 7), repeat=n_re):
                tot += score(kept + out)
            ev = tot / 6**n_re
        results.append((ev, kept))

results.sort(reverse=True)
for ev, kept in results[:5]:
    print(f"keep {kept}  EV = {ev:.4f}")

The script prints "keep $(4, 4, 4, 5, 5)$ EV $= 25.0000$" at the top, confirming the boxed strategy.

Riddler Classic

It is the final turn of a Yahtzee game and the only category left to score is a large straight (all five dice consecutive, so $1,2,3,4,5$ or $2,3,4,5,6$). On the first of the three rolls you roll $1, 2, 4, 5, X$, where $X \ne 3$. You may reroll the $X$ in hope of a $3$, or reroll the $1$ and the $X$ in hope of landing some combination of $3$ and $6$. What is the best strategy for hitting a large straight?

The Riddler, FiveThirtyEight, March 9, 2018(original post)

Solution

The setup. Two rolls remain. Each roll lets you choose a subset of dice to keep and reroll the rest. Success means ending the third roll with a large straight.

Strategy A: reroll only $X$. The hand becomes $(1, 2, 4, 5)$ plus one fresh die. Success requires that die to land on $3$. On the second roll this happens with probability $1/6$. If it fails, the same one-die reroll on the third roll has probability $1/6$ as well. So $$P_A = \frac{1}{6} + \frac{5}{6}\cdot\frac{1}{6} = \frac{11}{36} \approx 0.306.$$

Strategy B: reroll $\{1, X\}$. Hold $\{2, 4, 5\}$ and reroll two fresh dice. Now there are two routes to a large straight: pair $\{1, 3\}$ for $1\text{-}2\text{-}3\text{-}4\text{-}5$, or pair $\{3, 6\}$ for $2\text{-}3\text{-}4\text{-}5\text{-}6$. Treat the two new dice as an ordered pair, with $36$ equally likely outcomes, and split by what the second roll produced. In each branch we play the third roll optimally.

The counts close: $4 + 7 + 16 + 9 = 36$. Combining, $$\begin{align*} P_B &= \frac{4}{36}\cdot 1 + \frac{7}{36}\cdot\frac{1}{3} + \frac{16}{36}\cdot\frac{1}{6} + \frac{9}{36}\cdot\frac{1}{9} \\ &= \frac{24}{216} + \frac{14}{216} + \frac{16}{216} + \frac{6}{216} = \frac{60}{216} = \frac{10}{36} \approx 0.278. \end{align*}$$

Compare. $P_A = 11/36$ and $P_B = 10/36$. The single-die reroll wins by $1/36 \approx 2.8$ percentage points. $$\boxed{\,\text{Reroll only the } X.\quad P_A = \tfrac{11}{36}\, \text{vs.\ } P_B = \tfrac{10}{36}.\,}$$ A full dynamic-programming sweep over every subset of dice to keep (not only the two named strategies) gives the same optimal probability $11/36$ for any starting $X \ne 3$, including $X = 6$ (rerolling either the $1$ or the $X$ is equivalent: hold $\{2, 4, 5, 6\}$ and reroll the $1$ for a $3$).

The computation

Solve the Yahtzee large-straight subproblem by backward induction over the sorted-hand state with the number of rolls remaining.

1. Define $V(h, k)$ = probability of finishing with a large straight when the current sorted hand is $h$ and $k$ rolls remain.

2. At $k = 0$, return $1$ if $h$ is a large straight else $0$.

3. At $k > 0$, maximize over all subsets of $h$ to keep, averaging $V$ over the $6^{\text{rerolls}}$ outcomes.

from itertools import product, combinations
from functools import lru_cache

def is_LS(h):
    s = sorted(h)
    return s == [1,2,3,4,5] or s == [2,3,4,5,6]

@lru_cache(maxsize=None)
def V(h, k):
    if k == 0:
        return 1.0 if is_LS(h) else 0.0
    best = 0.0
    seen = set()
    for r in range(6):
        for keep_idx in combinations(range(5), r):
            kept = tuple(sorted(h[i] for i in keep_idx))
            if kept in seen: continue
            seen.add(kept)
            n_re = 5 - r
            if n_re == 0:
                v = 1.0 if is_LS(kept) else 0.0
            else:
                tot = 0.0
                for out in product(range(1,7), repeat=n_re):
                    tot += V(tuple(sorted(kept + out)), k - 1)
                v = tot / 6**n_re
            if v > best:
                best = v
    return best

for X in [1, 2, 4, 5, 6]:
    h = tuple(sorted((1, 2, 4, 5, X)))
    print(f"X={X}: optimal P = {V(h, 2):.6f} = 11/36 = {11/36:.6f}")

The script prints $11/36 \approx 0.305556$ for every $X \ne 3$, matching the boxed answer.